(i) All chords of the curve 3x^2 - y^2 - 2x + 4y = 0 which subtend right angle at the origin will pass through a fixed

Question

Passage: Suppose a line lx + my = 1 meets a seconddegree curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at two points A and B. Then, the combined equation of the pair of lines OA and OB is ax2 + 2hxy + by2 + (2gx + 2fy)(lx + my) + c(lx + my)2 = 0 which is a seconddegree homogeneous equation. Answer the following questions.

(i) All chords of the curve 3x2 - y2 - 2x + 4y= 0 which subtend right angle at the origin will pass through a fixed point whose coordinates are

(a) (-1,2)

(b) (1, - 2)

(c) (1,2)

(d) (-1,-2)

(ii) If the intercept of the line lx + my = 1 made by the curve x2+ y2 - a = 0 subtends right angle at the origin, then l2 + m2 is equal to

(a) 2/a2

(b) 1/a2

(c) 2a2

(d) 3a2

(iii) The line y = mx + c makes an intercept on the curve y - 4ax = 0 which subtends angle at the origin. Then, the line y = mx + c passes through a fixed point whose coordinates are

(a) (2a, 0)

(b) (a,0)

(c) (3a,0)

(d) (4a,0)

Answer 1

Correct option (i) (B) (ii) (A) (iii) (D)

Explanation

(i) Suppose lx + my = 1 is a line meeting the curve 3x2 - y2 - 2x + 4y = 0 at points A and B. Therefore, the combined equation of the pair of lines OA and OB is

3x2 - y2 -(2x - 4y)(lx + my) = 0

Since AOB = 90° , in the above equation, the coefficient of x2 + the coefficient of y2 = 0. Therefore

(3 - 2l) + (-1 + 4m) = 0

l - 2m - 1 = 0

Hence, the line lx + my - 1 = 0 passes through the point (1, -2)

(ii) Suppose the line lx + my = 1 intersects x2 + y2 = a2 at points A and B. Therefore, the combined equation of the pair of lines OA and OB is

x2 + y2 = a2(lx + my)2 = 0

Since ΔAOB = 90°, in the above equation, the coefficient of x2 + the coefficient of y2 = 0. That is,

(1 - a2l2) + (1 - a2m2) = 0

(iii) Suppose the line y = mx + c , where c ≠ 0. meets the curve y2 - 4ax = 0 at two points A and B. The combined equation of the pair of lines OA and OB is

Now, AOB = 90°⇒ the coefficient of x2 + the coefficient of y2 = 0. This mean

4am/c + 1 = 0

⇒ c = -4am

Therefore, the equation of the line is

y = mx + c = mx - 4am

⇒ y = m(x - 4a)

which passes through the fixed point (4a,0).